Question

If \( a \) is a rational number, then the roots of the equation \( x^2 - 3ax + a^2 - 2a - 4 = 0 \) are:

• A. rational and equal numbers
• B. different real numbers
• C. different rational numbers only
• D. not real numbers

Hint:

When calculating the discriminant of a quadratic equation, always check if the discriminant is non-negative to ensure that the roots are real. For distinct real roots, the discriminant must be strictly positive.

Answer 1

The Correct Option is B

We are given the quadratic equation:
\( x^2 - 3ax + a^2 - 2a - 4 = 0 \)
To determine the nature of the roots, we calculate the discriminant (\( \Delta \)) of the quadratic equation, which is given by:
\( \Delta = b^2 - 4ac \)
For the equation \( x^2 - 3ax + (a^2 - 2a - 4) = 0 \), the coefficients are:
\( a = 1, \, b = -3a, \, c = a^2 - 2a - 4 \)
Substituting into the discriminant formula:
\( \Delta = (-3a)^2 - 4(1)(a^2 - 2a - 4) = 9a^2 - 4(a^2 - 2a - 4) \)
Expanding and simplifying:
\( \Delta = 9a^2 - 4a^2 + 8a + 16 = 5a^2 + 8a + 16 \)
For the roots to be real, the discriminant must be non-negative:
\( \Delta = 5a^2 + 8a + 16 \geq 0 \)
Since \( 5a^2 + 8a + 16 \) is always positive for all real values of \( a \), the roots of the quadratic equation will always be real and distinct.
Thus, the correct answer is that the roots are “different real numbers.”